36 Comments

Jenna Page
on February 23, 2016 at 4:11 pm

I know we’ve already been taught this, but on the top of page 178 for the Proof, I’m confused how/what you plugged into the definition of a limit (how they got to the second line). Both the difference rule and the sum rule are easy to understand. For examples 1, 2, and 3 on page 186, why do you have to use the product rule and how did you know you had to use it? For example 4 on page 188, how did they get to the second line? For example 5, how do you know you have to use quotient rule?

Jonathan E Lopez
on February 25, 2016 at 11:08 am

For the sum/difference rule proof, think of the numerator as consisting of two parts:

x+h plugged into the function whose derivative you want – the original function.

Since we want the derivative of f(x)+g(x), the numerator becomes

f(x+h)+g(x+h) – (f(x)+g(x)

Note that this is x+h plugged into f+g minus f+g.

Generally if you have a term with more than one “x”, and there is multiplication, then the product rule will have to be used. For example, this is the case with f(x)=xe^x in example 1.

In example 2, they note that you can avoid using the product rule if you are willing to distribute (see solution 2).

In example 3, since f(x)=sqrt{x}g(x), there is more than one “x”, and multiplication, so product rule is needed.

For example 4, the quotient rule is used. The second line (that starts with y’=), they are indicating that they are doing BT’-TB’/B^2. In the next line, they actually calculate the derivatives T’ and B’. Then they simplify.

Example 5 requires the quotient rule because the function looks like a fraction and there is no way to simplify it. Note that the slash “/” means division, so the function is

y = e^x
_____
1+x^2

Mick Piombino
on February 23, 2016 at 5:50 pm

On page 177 in the example I’m kinda confused on how they got to the third line? it seems like they just changed equations? On page 188 the last example 5, did they just find slope? It doesn’t look like they did y2-y1=m(x2-x1)?

Jonathan E Lopez
on February 25, 2016 at 11:11 am

Which example are you looking at on page 177?

For example 5 on page 188, note that the slope is 0 (plug x=1 into the derivative). So that means the equation of the tangent line is

y-y1=0(x-x1), or just y-y1=0.

Sine the point is (1,.5e), this becomes y-.5e=0, or y=.5e.

Annika Smith
on February 23, 2016 at 6:15 pm

On page 185 I don’t understand their explanation of the product rule. I can somewhat see how they got a rectangle, but I don’t understand where the delta’s came from.

Jonathan E Lopez
on February 25, 2016 at 11:21 am

That proof is a bit trickier than the one we went over in class.

The delta’s mean “change”. So u and v are actually functions of x. If we start with the product uv (the white rectangle), and then change x by a little, then u and v (since they are functions of x) will change by a little. The amount by which u and v change are called delta(u) and delta(v). How much does the product change? Well, the product becomes

(u + delta(u))(v + delta(v)), which contains 4 terms if you expand (represented by the four rectangles in the picture on page 185). One of these terms is the original uv, and the other 3 terms are a result of the change in x that caused changes in u and v. So these three new terms are called the change in uv, or delta(uv).

If we divide this equation for delta(uv) by delta(x) and look at the limit as the size of the change gets very small, we obtain some derivatives and, ultimately, the product rule.

Shelby Gennocro
on February 23, 2016 at 6:43 pm

In example 5 line 3
why does d/dx(5) derive to zero? is it because it has no previous variable attached?

Jonathan E Lopez
on February 25, 2016 at 11:22 am

This is the first rule, “The Constant Rule”. The derivative of any constant is 0.

The reason is that derivatives measure change. If a function is constant, that means it does not change (or changes by 0). So the derivative is 0.

Samantha Caico
on February 23, 2016 at 9:46 pm

All the rules make sense. Sometimes I get lost with the the natural number e, like in page 186 example 1 I automatically see the x as a coefficient and not a separate function. The process for the derivative makes sense if I just try to separate the function into parts. Simplifying the derivative is the easy part, just making sense of the rules in various cases will be my main focus. The quotient rule threw me off a little because I don’t quite understand how the rule was derived with the g(x) being squared in the denominator and such, but I can follow the application of it with a quotient function without a problem once I tried to just memorize the format.

Jonathan E Lopez
on February 25, 2016 at 11:25 am

Right, recognizing when to use a particular rule is often more challenging than using the rule. In example 1, the x is technically the coefficient, but the problem is that it varies.

Think of any term that can not be simplified and that has more than one “x” as a term on which you’ll have to use one of the more complicated rules (the product rule, the quotient rule, or the chain rule (which we will see in section 3.4)).

The next extra credit problem will be related to the quotient rule, and if you try that you will see why we get a g(x)^2 in the denominator of the quotient rule.

Kianne Fernandez
on February 23, 2016 at 9:51 pm

I thought the rule for the product rule and quotient rule was keep, change +/- change keep? The book uses keep, change +/- keep, change. (used +/- because the product rule uses addition while the quotient rule uses subtraction)

Jonathan E Lopez
on February 25, 2016 at 11:27 am

Remember that multiplication is commutative (for example, 2*3=3*2). So the two things you wrote are actually the same, since

“change * keep” = “keep * change”.

I suggest remembering and using the rule in whatever way is most comfortable for you.

Alissa Buchta
on February 23, 2016 at 10:16 pm

On page 186 for example 1 I find it very confusing when you use the Product Rule more than one time and I find it hard to follow the steps of what exactly is going on in the problem. Example 2 is a bit easier to follow but I am still confused at to the steps that are taking place. Also on page 188 Example 5 deriving the slope of the tangent line and how exactly to get the answer was hard for me to understand.

Jonathan E Lopez
on February 25, 2016 at 11:32 am

In Example 1 on page 186, they are finding some higher derivatives.

To find the first derivative, use F=x and S=e^x. After simplifying, the derivative is f'(x)=(x+1)e^x.

Now we need to find the derivative of that function, which is written as a product of two things. Use F=x+1 and S=e^x. After simplifying, the derivative is f”(x)=(x+2)e^x. They then point out that there is a pattern developing.

In Example 2, the first line (starting with f'(t)=), they are “setting up” the product rule, by saying that they need to do F*S’+S*F’. In the next line, they actually fill in the derivatives S’ and F’. In class, we have been doing this part off to the side by explicitly listing F, S, and then F’, S’.

There is a comment above related to example 5. By plugging in x=1, the slope of the tangent line is 0. So that means the equation of the TL is y = y-int.

Jenivette Garcia
on February 24, 2016 at 10:12 am

The sum/difference rule are easy to understand and easy to solve because all you do is the derivative of each piece seperately. The product rule is also easy to follow on examples 1-3. It’s a little bit scary when there is a square root in the function, but the books walks you through it nicely. The proof the book gives of the quotient is very confusing I couldn’t follow it. Despite that, the quotient rule is also fairly manageable in terms of solving. The only thing I’m confused about is the dy/dx, d/dx symbols. What does the top and bottom suggest? I am used to just writing y’ or f'(x).

Jonathan E Lopez
on February 25, 2016 at 11:37 am

Yes, the proof of the quotient rule is not easy. There will be an extra credit problem that explores (an easier) version of the proof of the quotient rule.

If the function is written as y = …, then y’ or dy/dx are both symbols used to represent the derivative.

If the function is written as f(x) = …, then f'(x) is often used.

d/dx must be followed by a function in parentheses, as it means “the derivative of the function …”.

For example, if y = x^2, we can write

y’ = 2x (think of this as “the derivative of the thing we called y is 2x”)
dy/dx = 2x (think of this as “the derivative of the thing we called y is 2x”)
d/dx[x^2] = 2x (think of this as “the derivative of x^2 is 2x”).

If we have f(x) = x^2, we typically write

f'(x) = 2x, but could also use
d/dx[x^2] = 2x.

Antonio Roman
on February 24, 2016 at 10:38 am

The quotient rule and product rule allow you to break up the problem into smaller and easier pieces to solve. It can be difficult to see the simplest way to approach the problem. For example, problem 2 can be solved using the product rule, but it’s quicker to rewrite the equation. Practicing these types of problems is the only way to get used to seeing the quickest and simplest way to find the answer.

Jonathan E Lopez
on February 25, 2016 at 11:38 am

Yes, exactly. Practice is the only way to get really efficient at derivatives!

Benny Reese
on February 24, 2016 at 3:35 pm

I had taken a pre-calculus course in high school and this was the information that I remembered. For some reason this was the unit of material that stuck to me. But a question that can up when I was reading on page 188 is that why is it in example 4 that the denominator stays (x^3 + 6)^2. Like in class you did the same? Is it wrong for me to square the equation? But aside from that it was mostly a refresher for me.

Jonathan E Lopez
on February 25, 2016 at 11:40 am

When using the quotient rule, you generally leave the denominator in the form B^2, rather than actually squaring it.

B^2 is sort of like a “factored form”, which is more concise than if we were to square.

For example, (x^3+6)^2 is a bit nicer looking than x^6+12x^3+36, which is what you would get if you squared it. Either way is correct, though.

Melanie Rivera
on February 24, 2016 at 4:01 pm

The quotient, product, sum and difference all allow you to break apart the function and solve the smaller individual pieces separately which make it simpler to figure out. The power rule is easier to solve with whole integers rather than fractions as exponents due to the extra thought process but not too complicated. I’m a bit confused about the proof on page 177 because I don’t quite understand the constant multiple rule and how g(x) turns into f(x).

Jonathan E Lopez
on February 25, 2016 at 11:42 am

Right after the word proof, you see “let g(x)=cf(x)”. This means we start with a function f, and multiply it by c. This gives us a new function, and we can give that new function a name, such as g.

Since we want the derivative of cf(x), this is now the same as asking for the derivative of g(x). This allows us to write down the limit definition of the derivative.

Another way to think of the limit definition is like this: the numerator should be x+h plugged into the function whose derivative we want – the original function. In this case, that would be

cf(x+h) – cf(x)

You can then factor out a c, and get to where they are in the line 3 of the proof.

Emily Czechowski
on February 24, 2016 at 7:17 pm

As having taken calculus previously in high school the sum/difference rule is not difficult for me. The product and quotient rule were review for me as well, I find these tactics very easy. The examples on page 188 really helped illustrate the concepts more. Overall it was a nice refresher to read over the pages and the examples that were provided were easy to follow and understand.

Ben Eisenhut
on February 24, 2016 at 7:26 pm

Having taken calculus in high school, these pages were a good review. Although they are very similar in concept, The Quotient Rule always caused me trouble. After having gone over examples to illustrate both rules in class, and after reading this section, I feel like I have a good grasp on the material. One thing I found difficult was the form in which the problems were written. By having (d/dx) all over throughout the problem, I became confused more easily. So, I wrote out the problems with f'(x) and g'(x) and F and S (T and B) so that they would be easier to read.

Jonathan E Lopez
on February 25, 2016 at 11:44 am

That’s a good strategy. There is all sorts of notation that is used when working with derivatives, and the book really mixes it up. You should use whatever you are most comfortable with.

Therese Hetzer
on February 24, 2016 at 8:11 pm

I understood all of the rules and the process of solving the examples given, but I had a problem understanding solution 1 to example 2 on page 186. I understand solution 2 which is quicker and simpler, but where does the 3b come from in the solution? Why/how is it that 1/2t^-1/2 becomes 2 sqrt of t? Same for 1/2 x^-1/2 becoming 2 sqrt of x in example 3. Another small thing was the change from d/dx to dy/dx when it comes to the quotient rule. Is the dy the derivative of y= like I assume it is?

Jonathan E Lopez
on February 25, 2016 at 11:50 am

In solution 1 to example 2 on page 186, the second term is

(a+bt)*(1/2)t^{-1/2}.

This means there is an a+bt on top, a 2 on the bottom, and negative exponent means we can a t^{1/2} into the denominator. So this term simplifies to

a+bt
——-
2*sqrt(t)

Now they want to combine this with the first term, which is b*sqrt{t}. To do this, we need a common denominator. So b*sqrt(t) is multiplied by 2*sqrt(t)/2*sqrt(t), which results in

2bt
———-
2*sqrt(t)

When you add numerators, you get a+3bt.

In example 3, same idea. The term (1/2)x^{-1/2} already has a 2 in the denominator (because of the 1/2), and the power of x can be moved into denominator because of the negative exponent. The only thing left in the numerator is 1. So

1
—-
2x^{1/2}

As far as derivative notation, y’ and dy/dx actually stand for the derivative, while the d/dx notation means “the derivative of the function that follows”. For example, if y = x^2, we can write

y’ = 2x (think of this as “the derivative of the thing we called y is 2x”)
dy/dx = 2x (think of this as “the derivative of the thing we called y is 2x”)
d/dx[x^2] = 2x (think of this as “the derivative of x^2 is 2x”).

Anthony Rebmann
on February 24, 2016 at 8:19 pm

For the sum and difference rule basically tells us that we have to find the derivatives of each function, then add or subtract them. It doesn’t matter what order you do when your adding them both, but when finding the difference you have to follow the order as given due to the minus sign. As you were saying in class today for the product rule you have to take the first and second part of the equation. So the derivative of the first part times the the original second part, then the derivative of the second part times the original first part. Getting two terms then finding the sum of those two terms. The quotient rules works pretty much the same way except you find the difference between the two and divide by the original denominator squared (mind you have to keep it in the original order because the minus sign makes a difference.)

Jonathan E Lopez
on February 25, 2016 at 11:51 am

Yes, the order is definitely important in the Quotient Rule!

Alonda McEachin
on February 24, 2016 at 8:34 pm

All the rules up until the product/quotient rule I understand. I’m a bit confused on when I should opt for using the product rule vs the others. Is it in a case of simplification that I’ll know which rule to apply to a specific function or is there a trick to decide whether I apply the product/quotient rule. I’m a big fan of these rules though because the old process was much longer. The proof makes sense, I think it’s really a case of doing longhand to show that the new derivative is just the sum/difference of the two separate derivatives. The product/quotient rule is a bit taunting but in class the idea of “first”, “second”, “bottom”, and “top” makes the problem seem so much easier. Finding the derivative is so much easier than I thought it would be especially with these new rules.

Jonathan E Lopez
on February 25, 2016 at 11:53 am

If a function can be simplified, it may be possible to avoid using the product or quotient rules (for example, like in y = x(1+x^2)).

If you cannot simplify and a term involves more than one “x”, then you will have to use one of the more complicated rules (the product rule, quotient rule, or chain rule (which we cover in 3.4)).

Recognizing what rule to use is tricky, and is a skill that requires a lot of practice to develop.

Peter Bartnik
on February 24, 2016 at 9:46 pm

Would you be able to use these rules to determine instantaneous velocity with these rules? I understand that you can use the equation to prove that these rules work, but I wonder if there are any situations in which you can’t use these rules, or situations in which you shouldn’t use the rules.

Jonathan E Lopez
on February 25, 2016 at 11:56 am

Yes, instantaneous velocity is the derivative of position. So if you have a position function, and use the rules to find its derivative, then what you end up with is the velocity function (often the word “instantaneous” is dropped, but it is technically the “instantaneous velocity function”).

Yes, there are certainly cases where the rules don’t really apply. We typically focus more on using the rules, so generally deal with functions on which we can use them.

But the absolute value function is a good example. None of our rules will help us find the derivative of the absolute value function (whose derivative turns out to be discontinuous).

Marnisha Brooks
on February 24, 2016 at 11:37 pm

I understand the sum rule and how you use it to get the derivative of a function. When reading it I understood the proof example because you went over it in class so reading it was like review. On pages 186-187 example 1-3 was pretty straight forward to me and I like how they showed us what was happening step by step so i understood how to apply the product rule. On page 188 I understood how to apply the quotient rule but example five confused me because i don’t understand how they got rid of the 2x to get e^x(1-x)^2 in the numerator of the final answer.

Jonathan E Lopez
on February 25, 2016 at 11:58 am

In the numerator, they are basically factoring without telling you:

(1+x^2)e^x-e^x(2x)
= e^x(1+x^2-2x)
= e^x(x^2-2x+1)
= e^x(x-1)(x-1)
= e^x(x-1)^2
which is the same as e^x(1-x)^2.

Dan Emerson
on February 25, 2016 at 12:50 am

After taking pre-calculus and calculus in high school, I have a strong background with the concepts of the Sum/Difference Rules and the Product/Quotient Rules. It is nice to see the graphs to represent these concepts on pages 186 and 188. It is also nice to see how each rule actually works when solving it by using the limit definition of a derivative, that is new to me. Overall, I feel very comfortable with this topic.